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The x components of the average accelerations?

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  1. Oct 26, 2016 #1
    1. The problem statement, all variables and given/known data
    A(n) 8.5-kg object is sliding across the ice at 2.34 m/s in the positive x direction. An internal explosion occurs, splitting the object into two equal chunks and adding 12 J of kinetic energy to system. The explosive separation takes place over a 0.16-s time interval. Assume that the one of the chunks after explosion moves in the positive x direction. The x component of the average acceleration of this chunk during the explosion is afrontx, the x component fo the average acceleration of the other chunk during the explosion is arearx. What are the x components of the average accelerations of the two chunks during the explosion?

    2. Relevant equations
    k=1/2mv^2

    3. The attempt at a solution
    k=1/2(8.5)(2.34)^2
    = 23.27J

    23.27+12=35.27J
    35.27/0.16= 220.4m/s^2
    i've never done a problem like this
     
  2. jcsd
  3. Oct 26, 2016 #2
    What values are required to compute the average acceleration? ( or what's the formula for it).
     
  4. Oct 26, 2016 #3
    a=vf-vi/t
     
  5. Oct 26, 2016 #4
    Right. You've been given ##t## and the initial velocity for both the pieces. You need to find their final velocities. There are two unknowns, so we'll need 2 equations to solve them. Can you see what the equations can be?
     
  6. Oct 26, 2016 #5
    is one going to be the change in kinetic energy formula, and is the change momentum??
     
  7. Oct 26, 2016 #6
    is the initial velocity for both pieces just 2.34?
     
  8. Oct 26, 2016 #7
    Yes. You'll solve them to get the final velocities of both the pieces.
    That's right.
     
  9. Oct 26, 2016 #8
    for the mass do i use 8.5, or need to divide that by two since its split equally?
     
  10. Oct 26, 2016 #9
    When expressing the final momentum ( or kinetic energy) of the system, you'll have to divide it by two.
     
  11. Oct 26, 2016 #10
    but on the initial side, it would just be 8.5(2.34) for momentume, and 1/2(8.5)(2.34)^2 for kinetic energy?
     
  12. Oct 26, 2016 #11
    Yes, that's correct.
     
  13. Oct 26, 2016 #12
    i got answers of 20.48 and 25.16. they're wrong
     
  14. Oct 26, 2016 #13
  15. Oct 26, 2016 #14
    that was my work
     
  16. Oct 26, 2016 #15
    Your application of momentum conservation doesn't look right.
    Suppose ##v## to be the initial velocity of the block and ##v_1## and ##v_2## be the final velocities of the two pieces. If the mass of the block is ##m##, the conservation principle gives $$p_i = p_f $$
    $$ mv = \frac {m}{2}v_1 + \frac{m}{2}v_2 $$
    This is your first equation. Similarly, for kinetic energy you'll get $$ \frac{1}{2}mv^2 + 12 = \frac{1}{2}\frac{m}{2}(v_1)^2 +\frac{1}{2}\frac{m}{2}(v_2)^2 $$
    Solve these two equations for ##v_1##and ##v_2##. Use them in the formula for average acceleration.
     
  17. Oct 26, 2016 #16
    i got answers of 74.56 and -74.31 for acceleration which are wrong
     
  18. Oct 26, 2016 #17
    That doesn't match with my results either.
    What values did you get for ##v_1## and ##v_2## ? Show your work.
     
  19. Oct 26, 2016 #18
    IMG_7667.JPG
    i go values of 14.27 m/s and -9.59 m/s
     
  20. Oct 27, 2016 #19
    In the third line, you have substituted for ##v_2## in terms of ##v_1##. You seem to have made a mistake there.
    From the first equation(momentum conservation), we get ##2v = v_1 + v_2##. Or ##v_2 = 4.68 - v_1## . Use this in your second equation.
     
  21. Oct 27, 2016 #20
    i got accelerations of -117.136 m/s^2 and 117.136m/s^2
     
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